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y\'\'+4y\'+4y=e^-2x/x^3

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2006-01-17 12:27:52 · 1 個解答 · 發問者 龍斬澐 6 in 電腦與網際網路 軟體

1 個解答

1. y'' + 4y' + 4y = e-2x/x3sol:  特徵方程式( characteristic equation ):r2 + 4r + 4 = 0  → ( r + 2 )( r + 2 ) = 0 → r = - 2 , - 2 ~ 重根  yh = ( c1 + c2x )e-2x    = c1e-2x + c2xe-2x ~ 齊性解( homogeneous solution )  以參數變異法( method of variation of parameters )求特解:  令 yp = u1y1 + u2y2 = u1e-2x + u2xe-2x    Q = e-2x/x3  W( y1 , y2 ) =│ e-2x    xe-2x    │= e-4x        │ - 2e-2x e-2x - 2xe-2x │  u1 =∫( - y2Q/W )dx =∫{ [ - xe-2x‧( e-2x/x3 ) ]/e-4x }dx    =∫( - 1/x2 )dx = 1/x  u2 =∫( y1Q/W )dx =∫{ [ e-2x‧( e-2x/x3 ) ]/e-4x }dx    =∫( 1/x2 )dx = - 1/2x  yp = u1y1 + u2y2    = ( e-2x/x ) - ( e-2x/2x ) = e-2x/2x ~ 特解( particular solution )  通解( general solution ):y = yh + yp  → y = ( c1 + c2x )e-2x + ( e-2x/2x ) #

2006-01-17 15:32:46 · answer #1 · answered by 龍昊 7 · 0 0

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